The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.
Answers
We can study the choice of a reducing agent in a particular case using Ellingham diagram.
It is evident from the diagram that metals for which the standard free energy of formation oftheir oxides is more negative can reduce those metal oxides for which the standard free energy of formation of their respective oxides is less negative. It means that any metal will reduce the oxides of other metals which lie above it in the Ellingham diagram. This is because the standard free energy change (ΔrG°) of the combined redox reaction will be negative by an amount equal to the difference in Δf G° of the two metal oxides. Thus both Al and Zn can reduce FeO to Fe but Fe cannot reduce Al203 to A1 and ZnO to Zn. In the same way, G can reduce ZnO to Zn but not CO.
Note : Only that reagent will be preferred as reducing agent which will lead to decrease in free energy value (ΔG°) at a certain specific temperature.
MARK ME AS BRAINLIEST
Answer:
Answer with explanation-
I absolutely agree with the statement that choice of reducing agent depends on thermodynamic factors most importantly free energy change.
∆G° of the reducing agent must be more than the oxide it has to reduce. i.e. oxide with more negative ∆G° reduces oxides with less negative ∆G°
Examples-
(1) C (more negative ∆G°) can be used as reducing agent to reduce ZnO (less negative ∆G°).
ZnO + C --> Zn + CO
(2) Mg (more negative ∆G°) can be used as reducing agent to reduce Cu2O (less negative ∆G°).
Cu2O + Mg --> MgO + 2Cu